Solved lab

lab-sql-aggregation-and-transformation.sql

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+USE sakila;
+
+SHOW TABLES;
+
+-- answer to 1.1
+SELECT max(length) AS max_duration,
+	   min(length) AS min_duration
+FROM film;
+
+SELECT * FROM film;
+
+-- answer to 1.2
+SELECT AVG(length)
+FROM film;
+
+SELECT FLOOR(AVG(length)/60) AS hours,
+	   ROUND(MOD(AVG(length),60)) AS minutes
+FROM film;
+
+-- answer to 2.1
+SELECT DATEDIFF(
+	(SELECT MAX(rental_date) FROM rental),
+    (SELECT MIN(rental_date) FROM rental)) AS operating_days;
+
+-- answer 2.2
+SELECT *, DATE_FORMAT(rental_date, '%M') AS rental_month,
+	   DATE_FORMAT(rental_date, '%a') AS rental_day
+ FROM rental
+ LIMIT 20;
+
+-- answer 2.3
+SELECT *, 
+	CASE
+		WHEN DATE_FORMAT(rental_date, '%a') = "Sat" OR "Sun" THEN "weekend" 
+		ELSE "weekday"
+	END AS DAY_TYPE
+FROM rental;
+
+-- answer to 3
+SELECT title,
+	IFNULL(rental_duration, 'Not Available') AS adjusted_rental_duration
+FROM film
+ORDER BY title ASC;
+
+-- answer to 4
+SELECT CONCAT(first_name,' ',last_name) AS full_name,
+	   SUBSTRING(email, 1, 3) AS first_3_email
+FROM customer
+ORDER BY last_name ASC;
+
+-- Challenge 2
+-- answer to 1.1
+SELECT count(*) FROM film;
+
+-- answer to 1.2
+SELECT rating, count(film_id) AS number_of_films
+FROM film
+GROUP BY rating;
+
+-- answer 1.3
+SELECT rating, count(film_id) AS number_of_films
+FROM film
+GROUP BY rating
+ORDER BY number_of_films DESC;
+
+-- answer 2.1
+SELECT rating, ROUND(AVG(length),2) AS average_length 
+FROM film
+GROUP BY rating
+ORDER BY average_length DESC;
+
+-- answer 2.2
+SELECT rating, ROUND(AVG(length),2) AS average_length 
+FROM film
+GROUP BY rating
+HAVING average_length > 120
+ORDER BY average_length DESC;
+
+-- answer 3
+SELECT last_name
+FROM actor
+GROUP BY last_name
+HAVING COUNT(*) = 1;
+